Differential Equations(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

DIFFERENTIAL EQUATIONS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

The general solution of the differential equation $\frac{d y}{d x} = y t a n x - y^{2} s e c x$ is

tan x = (c + sec x)y
sec y = (c + tan y )x
sec x = (c + tan x)y
None of these

Discuss Question

Answer: Option C. -> sec x = (c + tan x)y
:
C
We have

\frac{d y}{d x} = y t a n x - y^{2} s e c x \Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} t a n x = - s e c x

Putting

\frac{1}{y} = v \Rightarrow \frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d v}{d x}

, we obtain

\frac{d v}{d x} + t a n x . v = s e c x

which is linear

I . F = e^{\int t a n x d x} = e^{l o g s e c x} = s e c x

∴

The solution is

v s e c x = \int s e c^{2} x d x + c \Rightarrow \frac{1}{y} s e c x = t a n x + c

\Rightarrow s e c x = y (c + t a n x)

Question 2.

If integrating factor of $x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0$ is $e^{\int P d x}$ , then P is equal to

$\frac{2 x^{2} - a x^{3}}{x (1 - x^{2})}$
$(2 x^{2} - 1)$
$\frac{2 x^{2} - 1}{a x^{3}}$
$\frac{(2 x^{2} - 1)}{x (1 - x^{2})}$

Discuss Question

Answer: Option D. ->

\frac{(2 x^{2} - 1)}{x (1 - x^{2})}

:
D

x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0 \frac{d y}{d x} + \frac{(2 x^{2} - 1)}{x (1 - x^{2})} y = \frac{a x^{2}}{(1 - x^{2})}, ∴ P = \frac{2 x^{2} - 1}{x (1 - x^{2})}

Question 3.

Solution of the equation $x d y = (y + x \frac{f (\frac{y}{x})}{f^{'} (\frac{y}{x})}) d x$

$f (\frac{x}{y}) = c y$
$f (\frac{y}{x}) = c x$
$f (\frac{y}{x}) = c x y$
$f (\frac{y}{x}) = 0$

Discuss Question

Answer: Option B. ->

f (\frac{y}{x}) = c x

:
B
We have,

x d y = (y + \frac{x f (\frac{y}{x})}{f^{'} (\frac{y}{x})}) d x

\Rightarrow \frac{d y}{d x} = \frac{y}{x} + \frac{f (\frac{y}{x})}{f^{'} (\frac{y}{x})}

which is homogenous
Putting

y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}

, we obtain

v + x \frac{d v}{d x} = v + \frac{f (v)}{f^{'} (v)} \Rightarrow \frac{f^{'} (v)}{f (v)} d v = \frac{d x}{x}

Integrating, we get log

f (v) = l o g x + l o g c

\Rightarrow l o g f (v) = l o g c x \Rightarrow f (\frac{y}{x}) = c x

Question 4.

The solution of the differential equation $(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0$ , is

$2 x e^{t a n^{- 1} y}, = e^{2 t a n^{- 1} y} + k$
$x e^{t a n^{- 1} y}, = e^{t a n^{- 1} y} + k$
$x e^{2 t a n^{- 1} y}, = e^{- t a n^{- 1} y} + k$
$(x - 2) k e^{t a n^{- 1} y}$

Discuss Question

Answer: Option A. ->

2 x e^{t a n^{- 1} y}, = e^{2 t a n^{- 1} y} + k

:
A

\frac{d x}{d y} + \frac{1}{1 + y^{2}} x = \frac{1}{1 + y^{2}} e^{t a n^{- 1} y}

I . F = e^{\int \frac{1}{1 + y^{2}} d y} = e^{t a n^{- 1} y}

∴

Solution is

x . e^{t a n^{- 1}} y

= \int e^{t a n^{- 1} y} . \frac{1}{1 + y^{2}} e^{t a n^{- 1}} d y = \frac{1}{2} e^{2 t a n^{- 1} y} + \frac{1}{2} k \Rightarrow 2 x e^{t a n^{- 1} y} = e^{2 t a n^{- 1} y} + k

Question 5.

The orthogonal trajectories of the family of curves $a^{n - 1} y = x^{n}$ are given by

$x^{n} + n^{2} y$ =constant
$n y^{2} + x^{2}$ =constant
$n^{2} x + y^{n}$ =constant
$n^{2} x - y^{n}$ =constant

Discuss Question

Answer: Option B. ->

n y^{2} + x^{2}

=constant

:
B
Differentiating, we have

a^{n - 1} \frac{d y}{d x} = n x^{n - 1} \Rightarrow a^{n - 1} = n x^{n - 1} \frac{d x}{d y}

Putting this value in the given equation, we have

n x^{n - 1} \frac{d x}{d y} y = x^{n}

Replacing

\frac{d y}{d x}

- \frac{d y}{d y}

, we have

n y = - x \frac{d x}{d y}

\Rightarrow n y d y + x d x = 0 \Rightarrow n y^{2} + x^{2}

=constant. Which is the required family of orthogonal trajectories.

Question 6.

The solution of $(y (1 + x^{- 1}) + s i n y) d x + (x + l o g x + x c o s y) d y = 0$ is

$(1 + y^{- 1} s i n y) + x^{- 1} l o g x = C$
$(y + s i n y) + x y l o g x = C$
$x y + y l o g x + x s i n y = C$
$N o n e o f t h e s e$

Discuss Question

Answer: Option C. ->

x y + y l o g x + x s i n y = C

:
C
The given equation can be written as

y (1 + x^{- 1}) d x + (x + l o g x) d y + s i n y d x + x c o s y d y = 0

\Rightarrow d (y (x + l o g x)) + d (x s i n y) = 0 \to y (x + l o g x) + x s i n y = C

Question 7.

The solution of $y_{2} - 7 y_{1} + 12 y = 0$ is

$y = C_{1} e^{3 x} + C_{2} e^{4 x}$
$y = C_{1} x e^{3 x} + C_{2} e^{4 x}$
$y = C_{1} e^{3 x} + C_{2} x e^{4 x}$
$N o n e o f t h e s e$

Discuss Question

Answer: Option A. ->

y = C_{1} e^{3 x} + C_{2} e^{4 x}

:
A
The given equation can be written as

(\frac{d}{d x} - 3) (\frac{d y}{d x} - 4 y) = 0 . . . . (i)

\frac{d y}{d x} - 4 y = u

then (I) reduces to

\frac{d u}{d x} - 3 u = 0

\Rightarrow \frac{d u}{u} = 3 d x \Rightarrow u = C_{1} e^{3 x}

. Therefore, we have

\frac{d y}{d x} - 4 y = C_{1} e^{3 x}

which is a linear equation whose I.F.is

e^{- 4 x}

. So

\frac{d}{d x} (y e^{- 4 x}) = C_{1} e^{- x}

\Rightarrow y e^{- 4 x} = - C_{1} e^{- x} + C_{2} \Rightarrow y = C_{1} e^{3 x} + C_{2} e^{4 x}

Question 8.

The solution lof $\frac{d y}{d x} - x t a n (y - x) = 1$

$c e \frac{e^{x^{2}}}{2} = s i n (y - x)$
$c e \frac{e^{x^{2}}}{2} = s i n (y + x)$
$c e \frac{e^{x^{2}}}{2} = s i n (y - x)^{2}$
$c e \frac{e^{x^{2}}}{2} = c o s (y - x)$

Discuss Question

Answer: Option A. ->

c e \frac{e^{x^{2}}}{2} = s i n (y - x)

:
A
Put y-x = z. Then

\frac{d y}{d x} - 1 = \frac{d z}{d x} \Rightarrow \frac{d y}{d x} = 1 + \frac{d z}{d x}

Given

\frac{d y}{d x} - x t a n (y - x) = 1 \Rightarrow 1 + \frac{d z}{d x} - x t a n z = 1 \Rightarrow \frac{d z}{d x} = x t a n z \Rightarrow \frac{1}{t a n z} d z = x d x \Rightarrow \int c o t z d z = \int x d x

\Rightarrow l o g | s i n z | - l o g c = \frac{x^{2}}{2} \Rightarrow l o g | \frac{s i n z}{c} | = \frac{x^{2}}{2} = \frac{s i n z}{c} = \frac{e^{x^{2}}}{2} \Rightarrow s i n (y - x) = c \frac{e^{x^{2}}}{2}

∴

The solution is

s i n (y - x) = c \frac{e^{x^{2}}}{2}

, where c is arbitrary constant.

Question 9.

Solution of the differential equation : $\frac{d y}{d x} = \frac{3 x^{2} y^{4} + 2 x y}{x^{2} - 2 x^{3} y^{3}}$ is

$x^{2} y^{2} + \frac{x^{2}}{y} = c$
$x^{3} y^{2} + \frac{x^{2}}{y} = c$
$x^{3} y^{2} + \frac{y^{2}}{x} = c$
$x^{2} y^{3} + \frac{x^{2}}{y} = c$

Discuss Question

Answer: Option B. ->

x^{3} y^{2} + \frac{x^{2}}{y} = c

:
B

x^{2} d y - 2 x^{3} y^{3} d y = 3 x^{2} y^{4} d x + 2 x y d x \Rightarrow x^{2} d y - 2 x y d x = 3 x^{2} y^{4} d x + 2 x^{3} y^{3} d y \Rightarrow \frac{2 x y d x - x^{2} d y}{y^{2}} + 3 x^{2} y^{2} d x + 2 x^{3} y d y = 0 \Rightarrow d (\frac{x^{2}}{y}) + d (x^{3} y^{2}) = 0 \Rightarrow \frac{x^{2}}{y} + x^{3} y^{2} = C

Question 10.

The degree of the differential equation satisfying the relation $\sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} = λ (x \sqrt{1 + y^{2}} - y \sqrt{1 + x^{2}})$ is

1
2
3
none of these

Discuss Question

Answer: Option A. -> 1
:
A
On Putting

x = t a n A, y = t a n B

we get

s e c A + s e c B = λ (t a n A s e c B - t a n B s e c A) c o s A + c o s B = λ (s i n A - s i n B) t a n (\frac{A - B}{2}) = \frac{1}{λ} t a n^{- 1} x - t a n^{- 1} y = 2 t a n^{- 1} \frac{1}{λ}

On differentiating

\frac{1}{1 + x^{2}} - \frac{1}{1 + y^{2}} \frac{d y}{d x} = 0

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